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1 year ago

7

Justine is ice-skating at the Lloyd Center what is her final velocity if she accelerates at a rate of 2.0 meters per second for

3.5 seconds

1 answer:

lilavasa [31]1 year ago

7 0

2*3.5 = 7m/s

You multiply the acceleration per the time (they both are in seconds, otherwise, you should set them in the same units).

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A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the

kodGreya [7K]

Answer:

3.73994

No,

Explanation:

$P_1$ = Pressure at the bottom of the lake = 3.5 atm

$P_2$ = Pressure at the top of the lake = 1 atm

$V_b$ = Volume at the bottom of the lake

$V_s$ = Volume at the top of the lake

$T_1$ = Temperature at the bottom of the lake = 4 °C

$T_2$ = Temperature at the top of the lake = 23 °C

From ideal gas law we have the relation

$\dfrac{P_1V_b}{T_1}=\frac{P_2V_s}{T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{P_2T_1}{P_1T_2}\\\Rightarrow \dfrac{V_b}{V_s}=\frac{1\times 277.15}{3.5\times 296.15}\\\Rightarrow \dfrac{V_s}{V_b}=0.26738^{-1}\\\Rightarrow \dfrac{V_s}{V_b}=3.73994$

The ratio is 3.73994

As Jacques is ascending if he holds his breath his lungs acting like a bubble would expand. Hence, it is not safe to hold his breath while ascending,

8 0

2 years ago

A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s east find the vel

Harlamova29_29 [7]

Answer:

38.0 m/s east

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.311) (30.3) + (0.0570) (19.2) = (0.311) v₁ + (0.0570) v₂

10.52 = 0.311 v₁ + 0.0570 v₂

In an elastic collision, kinetic energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.311) (30.3)² + (0.0570) (19.2)² = (0.311) v₁² + (0.0570) v₂²

306.5 = 0.311 v₁² + 0.0570 v₂²

Solve the system of equations.

0.311 v₁ = 10.52 − 0.0570 v₂

v₁ = 33.82 − 0.1833 v₂

306.5 = 0.311 (33.82 − 0.1833 v₂)² + 0.0570 v₂²

306.5 = 0.311 (1144 − 12.40 v₂ + 0.03360 v₂²) + 0.0570 v₂²

306.5 = 355.7 − 3.856 v₂ + 0.01045 v₂² + 0.0570 v₂²

0 = 0.06745 v₂² − 3.856 v₂ + 49.16

Use quadratic formula.

v₂ = [ 3.856 ± √(14.87 − 13.26) ] / 0.1349

v₂ = 19.2 or 38.0

We know v₂ isn't 19.2 m/s, so v₂ = 38.0 m/s.

6 0

1 year ago

Bill has a mass of 85 kg and is skating west. He increases his speed from 3 m/s to 5 m/s by applying a force for 3 seconds. What

likoan [24]

F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N

4 0

2 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

2 years ago

Identify whether the features and functions belong to the Hubble Space Telescope or the ALMA radio telescope. 1.installed on lan

tester [92]

Hublle Space Telescope
- Orbits around Earth.
- Receive light before it enters Earth's Atmosphere.
- Get information about stars before they are born.

Alma Radio Telescope
- Installed on land.
- Receive light after it passes through Earth's Atmosphere.
- Get information about stars after they are born.

5 0

2 years ago

Read 2 more answers